Molal depression constant for a solvent is 4.0 K kg mol$$^{-1}$$. The depression in the freezing point of the solvent for 0.03 mol kg$$^{-1}$$ solution of K$$_2$$SO$$_4$$ is: (Assume complete dissociation of the electrolyte)

We recall that the lowering of the freezing point for a solution is given by the relation of colligative properties. The formula is stated first:

$$\Delta T_f = i \, K_f \, m$$

where

$$\Delta T_f$$ = depression in freezing point (in kelvin),
$$i$$ = van’t Hoff factor (accounts for the number of particles produced by dissociation),
$$K_f$$ = molal depression constant of the solvent (in K kg mol−1),
$$m$$ = molality of the solution (in mol kg−1).

We are given that the solvent has $$K_f = 4.0\ \text{K kg mol}^{-1}$$ and the solution has a molality $$m = 0.03\ \text{mol kg}^{-1}$$.

Now, the solute is potassium sulfate, $$\text{K}_2\text{SO}_4$$. On complete dissociation, it produces two potassium ions and one sulfate ion:

$$\text{K}_2\text{SO}_4 \;\; \longrightarrow \;\; 2\,\text{K}^+ \;+\; \text{SO}_4^{2-}$$

Thus, the total number of ions formed per formula unit is

$$2 + 1 = 3$$

Hence, for complete dissociation, the van’t Hoff factor is

$$i = 3$$

Substituting everything into the freezing-point depression formula, we have

$$\Delta T_f = i \, K_f \, m = 3 \times 4.0\ \text{K kg mol}^{-1} \times 0.03\ \text{mol kg}^{-1}$$

Carrying out the multiplication step by step:

First multiply $$4.0$$ by $$0.03$$:

$$4.0 \times 0.03 = 0.12$$

Now multiply by the van’t Hoff factor $$3$$:

$$3 \times 0.12 = 0.36$$

So, we obtain

$$\Delta T_f = 0.36\ \text{K}$$

Hence, the correct answer is Option A.